#### Question

Find the frequency of revolution of an electron in Bohr’s 2nd orbit; if the radius and speed of electron in that orbit is 2.14 × 10^{-10} m and 1.09 × 10^{6} m/s respectively. [π= 3.142]

#### Solution 1

Given

r_{2} = 2.14 × 10^{-10} m

n = 2

v_{2} = 1.09 × 10^{6} m/s

To find: Frequency of revolution (`nu_2`)

`v = romega=r(2pinu)`

`nu=v/(2pir)`

`nu_2=v_2/(2pir_2)=(1.09xx10^6)/(2xx3.142xx2.14xx10^-10)`

`nu_2=8.11xx10^14 Hz`

The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 10^{14} Hz.

#### Solution 2

`T = (2pir)/V`

`because T=1/f`

`f = V/(2pir)`

`f=(1.09 xx 10^6)/(2 xx 3.14 xx 2.14 xx 10^-10)`

`f=8.11 xx 10^14 Hz`

The frequency of revolution of electron in 2nd Bohr orbit is 8.11 × 10^{14} Hz.

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#### APPEARS IN

Solution Find the Frequency of Revolution of an Electron in Bohr’S 2nd Orbit; If the Radius and Speed of Electron in that Orbit is 2.14 x 10^-10 M and 1.09 x 10^6 M/S Respectively. Concept: Bohr'S Model for Hydrogen Atom.